The Chief Examiner reported that though the question was attempted by majority of the candidates, their performance was very poor. Majority of the candidates were reported not to resolve the given forces correctly and as such could not find the required angle. Candidates were expected to show that by resolving the forces vertically, we obtain:
(10 cos 30o) N + (8cos 45o)N – (20 sin 60o)N = 5 N + 8 N – 10N 
= (8 5 )N = - 0.66N.
Resolving the forces horizontally gave:
(8o) N  (10 sin30o)N (20 cos 60o)N = (8 -5 -10)N = 7N. Candidates would thereafter show that if θ was the required angle, then tan θ =   Therefore tan -1 (0.0943) = 5.39o. Since  was in the third quadrant (both the numerator and denominator is negative), it implied that the required direction was = (270 – 5.39)o  = 265o or S84.61oW.